(P+frac{a}{V^2})(V-b)=RT represents the equat | vedclass

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(C) According to the principle of homogeneity of dimensions,the dimensions of terms added or subtracted must be the same.Since $b$ is subtracted from

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Compressibility

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(C) According to the principle of homogeneity of dimensions,the dimensions of terms added or subtracted must be the same.Since $b$ is subtracted from $V$,the dimensions of $b$ are equal to the dimensions of volume $V$:$[b] = [V] = [L^3]$Similarly,$\frac{a}{V^2}$ is added to $P$,so the dimensions of $\frac{a}{V^2}$ are equal to the dimensions of pressure $P$:$[\frac{a}{V^2}] = [P] \implies [a] = [P][V^2] = [ML^{-1}T^{-2}][L^6] = [ML^5T^{-2}]$Now,we find the dimensions of $\frac{b^2}{a}$:$[\frac{b^2}{a}] = \frac{[L^3]^2}{[ML^5T^{-2}]} = \frac{[L^6]}{[ML^5T^{-2}]} = [M^{-1}LT^2]$We know that compressibility $K$ is the reciprocal of the bulk modulus $B$:$[K] = \frac{1}{[B]} = \frac{1}{[ML^{-1}T^{-2}]} = [M^{-1}LT^2]$Thus,the dimensional formula of $\frac{b^2}{a}$ is the same as that of compressibility.

$(P+\frac{a}{V^2})(V-b)=RT$ represents the equation of state of some gases. Where $P$ is the pressure,$V$ is the volume,$T$ is the temperature and $a, b, R$ are the constants. The physical quantity,which has the same dimensional formula as that of $\frac{b^2}{a}$,will be

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